Answer
$$\frac{{{\partial ^2}z}}{{\partial {t^2}}} = - {\omega ^2}{c^2}\sin \omega x\sin \omega ct = {c^2}\underbrace {\left[ { - {\omega ^2}\sin \omega x\sin \omega ct} \right]}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}}$$
Work Step by Step
$$\eqalign{
& z = \sin \omega ct\sin \omega x \cr
& {\text{Find }}\frac{{{\partial ^2}z}}{{\partial {t^2}}}{\text{ and }}\frac{{{\partial ^2}z}}{{\partial {x^2}}} \cr
& \frac{{\partial z}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {\sin \omega ct\sin \omega x} \right] = \sin \omega x\frac{\partial }{{\partial t}}\left[ {\sin \omega ct} \right] = \sin \omega x\left( {\omega c\cos \omega ct} \right) \cr
& \frac{{\partial z}}{{\partial t}} = \omega c\sin \omega x\cos \omega ct \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = \frac{\partial }{{\partial t}}\left[ {\omega c\sin \omega x\cos \omega ct} \right] = \omega c\sin \omega x\frac{\partial }{{\partial t}}\left[ {\cos \omega ct} \right] \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = \omega c\sin \omega x\left( { - \omega c\sin \omega ct} \right) = - {\omega ^2}{c^2}\sin \omega x\sin \omega ct \cr
& and \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\sin \omega ct\sin \omega x} \right] = \sin \omega ct\frac{\partial }{{\partial x}}\left[ {\sin \omega x} \right] \cr
& \frac{{\partial z}}{{\partial x}} = \omega \sin \omega ct\cos \omega x \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ {\omega \sin \omega ct\cos \omega x} \right] = \omega \sin \omega ct\frac{\partial }{{\partial x}}\left[ {\cos \omega x} \right] \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \omega \sin \omega ct\left( { - \omega \sin \omega x} \right) = - {\omega ^2}\sin \omega x\sin \omega ct \cr
& {\text{Therefore,}} \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = - {\omega ^2}{c^2}\sin \omega x\sin \omega ct = {c^2}\underbrace {\left[ { - {\omega ^2}\sin \omega x\sin \omega ct} \right]}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}} \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}z}}{{\partial {x^2}}} \cr} $$
